An integral that equals its own mirror image

Step 1: substitute. With $x=\tan\theta$, $dx=(1+\tan^2\theta)\,d\theta$ and $\theta$ runs from $0$ to $\frac\pi4$:
$$I=\int_0^{\pi/4}\ln(1+\tan\theta)\,d\theta.$$

Step 2: the mirror. Replace $\theta$ by $\frac\pi4-\theta$:
$$I=\int_0^{\pi/4}\ln\!\left(1+\frac{1-\tan\theta}{1+\tan\theta}\right)d\theta=\int_0^{\pi/4}\ln\frac2{1+\tan\theta}\,d\theta=\frac\pi4\ln2-I.$$

Step 3: conclude. $2I=\frac\pi4\ln2$, so
$$I=\frac{\pi\ln2}8.$$

Numerical check. $\frac{\pi\ln2}8\approx0.27220$, matching a direct numerical integration.

Remark. The integral looks impossible to do by antiderivatives (it is not elementary). The trick is to add the integral to its own reflection and let the logarithms combine.